How Does Fission Energy Relate to the Semi-Empirical Mass Formula?

In summary, the problem at hand involves using the semi-empirical mass formula to prove that when an even-even nucleus with a large Z decays into two identical odd-even fragments, energy will be liberated if the condition c1(A) + c2(Z^2) + c3(A)^(-5/12) > 0 is satisfied. The Z and A values of each fragment will be 1/2 of the original nucleus. This condition is derived from the fact that for energy to be released, the binding energy of the initial nucleus must be greater than two times the binding energy of one of the resultant nuclei. The odd-even nature of the fragments means that the last term of the semi-empirical mass formula is
  • #1
hhhmortal
176
0

Homework Statement



Use the semi-empirical mass formula to show that when an even-even nucleus of large Z undergoes fission into two identical odd-even fragments energy will be liberated provided that the approximate condition:

c1(A) + c2(Z²) + C3(A)^-5/12 > 0

is satisfied and give values for the constants c1, c2 and c3.



Homework Equations



Semi-empirical mass formula


The Attempt at a Solution



Do I use an example of any atom which is even-even and two other atoms which are odd-even. Then put it into the semi-empirical mass formula and then what exactly do I need?
 
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  • #2
^Bump
 
  • #3
No, you need to demonstrate that it's true for any even-even nucleus of large Z and A decaying into two identical odd-even fragments (so what would the Z and A values of each fragment be?).
 
  • #4
Avodyne said:
No, you need to demonstrate that it's true for any even-even nucleus of large Z and A decaying into two identical odd-even fragments (so what would the Z and A values of each fragment be?).

The N = odd and Z= even for both identical fragments. But how would this help me?
 
  • #5
I'm having problems with the same question.

My thinking so far:

Each of the new nuclei will have 1/2A and 1/2Z

For energy to be released the binding energy of the intial nuclei must be greater than two times the binding energy for one of the resultant nuclei.

The odd-even means that the last term of the semi-empirical mass formula is 0, for even-even if delta(Z,A) is 1

I've tried setting this up as an inequality and rearranging but I can't see how an A^(-5/12) would be arrived at. Are there terms that can be neglected for a large Z or was that just in the question to show the SEMF would be accurate in this case?
 

Related to How Does Fission Energy Relate to the Semi-Empirical Mass Formula?

What is the Semi-empirical Mass Formula?

The Semi-empirical Mass Formula is a mathematical equation used to predict the approximate nuclear binding energy of an atomic nucleus based on the number of protons and neutrons it contains.

How does the Semi-empirical Mass Formula work?

The formula takes into account the contributions of the strong nuclear force, the Coulomb force, and the asymmetry term in the nucleus to calculate the binding energy. It also considers the fact that the mass of a nucleus is slightly less than the combined mass of its individual nucleons.

Why is it called "semi-empirical"?

The term "semi-empirical" refers to the fact that the formula is based on both theoretical calculations and experimental data. The theoretical part of the formula is based on the liquid drop model of the nucleus, while the empirical part is based on experimental measurements of nuclear binding energies.

What is the significance of the Semi-empirical Mass Formula?

The Semi-empirical Mass Formula is important because it allows scientists to predict the stability of a nucleus and understand the processes involved in nuclear reactions. It also provides a framework for studying nuclear properties and developing new models for the nucleus.

Are there any limitations to the Semi-empirical Mass Formula?

While the Semi-empirical Mass Formula is a useful tool, it does have limitations. It is most accurate for stable nuclei with even numbers of protons and neutrons, and it does not account for the effects of nuclear spin or the distribution of charge within the nucleus. It also cannot predict the exact binding energies of nuclei, but rather gives an approximation.

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